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36                    Finite Element Modeling and Simulation with ANSYS Workbench



                 EXAMPLE 2.3

                                             3
                                                 45°
                                                   2   P 2


                                              Y        2   P 1
                                                    1
                                                  45°
                                                      X
                                             1

                 A simple plane truss is made of two identical bars (with E, A, and L), and loaded as
                 shown in the above figure.
                  Find
                     a.  displacement of node 2;
                     b.  stress in each bar.

                 Solution
                 This simple structure is used here to demonstrate the FEA procedure using the bar ele-
                 ment in 2-D space. In local coordinate systems, we have

                                              EA    1  − 1
                                           ′
                                          k 1 =          =  k 2 ′
                                               L   − 1  1  
                  These two matrices cannot be assembled together, because they are in different coor-
                 dinate systems. We need to convert them into global coordinate system OXY.
                  Element 1:

                                                o
                                           θ= 45 , l  = m  =  2
                                                          2
                  Using Equation 2.32 or 2.33, we obtain the stiffness matrix in the global system

                                                    u 1  v 1  u 2  v 2
                                                     1  1  − 1  − 1
                                                   
                                                EA 1    1   − 1  − 1   
                                                   
                                           ′
                                         T
                                     1 k =  1 T kT =
                                           11
                                                2 L −   1  − 1  1  1 
                                                     1  − 1  1   
                                                    −          1  
                  Element 2:
                                                      2       2
                                             o
                                       θ= 135 , l  =−  , m  =
                                                     2        2
                                                    u 2  v 2  u 3  v 3
                                                     1  − 1  − 1  1 
                                                   
                                                EA − 1  1    1   − 1   
                                                   
                                         T
                                           ′
                                     2 k =  2 T kT =
                                           2
                                             2
                                                2 L −   1  1  1  − 1
                                                       −   −      
                                                      1  1  1  1  
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